Find $\frac{\partial f}{\partial x}$ where $f=h(x+y,y^2,x+z)$
I think that the solution is $(1,0,1)$, but I am not sure why that is the case. How am I to compute a partial derivative without a definition for the function f? I know the chain rule, but I don't know why $\frac{\partial f}{\partial u}$ where $u=x+y$ is $(1,0,0)$, as seemed to be the case when my professor did this example.
You have two functions. First $$m:\mathbb{R}^{3}\to\mathbb{R}^{3}\\(x,y,z)\mapsto(x+y,y^2,x+z)$$ And then another function $$h:\mathbb{R}^3\to\mathbb{R}$$ for which we don't have an explicit prescription.
Your function $f:\mathbb{R}^{3}\to\mathbb{R}$ is defined as the composition $h\circ m$. Then, by the chainrule, we have $$\frac{\partial f(x,y,z)}{\partial x} = \sum_{i=1}^{3} \frac{\partial h(a_1,a_2,a_3)}{\partial a_{i}} \Bigg|_{(a_1,a_2,a_3)=m(x,y,z)} \frac{\partial m^{i}(x,y,z)}{\partial x}$$
See if you can continue from there.